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1 year ago

Please help me! I am very confused

Mathematics

1 answer:

Sphinxa [80]1 year ago

4 0

Answer:

1) 22.5 A

2) 112 B

3) 430 L

4) 576 P

5) 486 L

6) 624 S

Code is A B L P L S

Step-by-step explanation:

1) Area of a triangle = 1/2 * base * height = 1/2 *5*9 = 22.5

2) Area of a parallelogram = base x height = 14 x 8 = 112

3) Area of a rectangular prism = 2(length * width) + 2*(length + height) + 2 *(length *height)

= 2(15x7 + 15*5 + 7*5)

= 2(105 + 75 + 35)

= 2 * 215

= 430

4) Volume of a triangular prism = 1/2 * base * length * height

= 1/2 * 8 * 16 * 9

= 576

5) A cube has six surfaces. Each surface has an area of s x s where s is the length of each side. In this case, each side has area of 9x9 = 81. Total surface area = 6 x 81 = 486 and that is the paper required

6) The trailer is a rectangular prism so its volume = length x width x height = 13 6 x 8 = 624

Now you have to look at each value and see which letter it corresponds to. For example answer 1) is 22.5 which lies between 0-100 so it gets letter A, answer (2) is 112 which lies in the range 101-200 so it gets the letter B and so on

You might be interested in

Which of the following statements is true for all sets A,B and C? Give a proof

ELEN [110]

Answer:

(a) and (b) are not true in general. Refer to the explanations below for counterexamples.

It can be shown that (c) is indeed true.

Step-by-step explanation:

This explanation will use a lot of empty sets $\phi$ just to keep the counterexamples simple.

Note that $A \cap B$ can well be smaller than $A$ . It should be alarming that the question is claiming $A\!$ to be a subset of something that can be smaller than $\! A$ . Here's a counterexample that dramatize this observation:

Consider:

$A = \left\lbrace 1 \right\rbrace$ .
$B = \phi$ (an empty set, same as $\left\lbrace \right\rbrace$ .)
$C = \phi$ (another empty set.)

The intersection of an empty set with another set should still be an empty set:

$A \cap B = \left\lbrace 1\right\rbrace \cap \left\lbrace\right\rbrace = \left\lbrace\right\rbrace$ .

The union of two empty sets should also be an empty set:

$((A \cap B) \cup C) = \left\lbrace\right\rbrace \cup \left\lbrace\right\rbrace = \left\lbrace\right\rbrace$ .

Apparently, the one-element set $A = \left\lbrace 1 \right\rbrace$ isn't a subset of an empty set. $A \not \subseteq ((A\cap B) \cup C)$ . Contradiction.

Consider the same counterexample

$A = \left\lbrace 1 \right\rbrace$ .
$B = \phi$ (an empty set, same as $\left\lbrace \right\rbrace$ .)
$C = \left\lbrace 2 \right\rbrace$ (another empty set.)

Left-hand side:

$(A \cup B) \cap C = \left(\left\lbrace 1 \right\rbrace \cup \left\lbrace \right\rbrace\right) \cap \left\lbrace 2 \right\rbrace\right = \left\lbrace 1 \right\rbrace \cap \left\lbrace 2 \right\rbrace = \left\lbrace \right\rbrace$ .

Right-hand side:

$(A \cap B) \cup C = \left(\left\lbrace 1 \right\rbrace \cap \left\lbrace \right\rbrace\right) \cup \left\lbrace 2 \right\rbrace\right = \left\lbrace \right\rbrace \cup \left\lbrace 2 \right\rbrace = \left\lbrace 2 \right\rbrace$ .

Apparently, the empty set on the left-hand side $\left\lbrace \right\rbrace$ is not the same as the $\left\lbrace 2 \right\rbrace$ on the right-hand side. Contradiction.

Part one: show that left-hand side is a subset of the right-hand side.

Let $x$ be a member of the set on the left-hand side.

$x \in (A \backslash B) \cap C$ .

$\implies x\in A \backslash B$ and $x \in C$ (the right arrow here reads "implies".)

$\implies x \in A$ and $x \not\in B$ and $x \in C$ .

$\implies (x \in A\cap C)$ and $x \not\in B \cap C$ .

$\implies x \in (A \cap C) \backslash (B \cap C)$ .

Note that $x \in (A \backslash B) \cap C$ (set on the left-hand side) implies that $x \in (A \cap C) \backslash (B \cap C)$ (set on the right-hand side.)

Therefore:

$(A \backslash B) \cap C \subseteq (A \cap C) \backslash (B \cap C)$ .

Part two: show that the right-hand side is a subset of the left-hand side. This part is slightly more involved than the first part.

Let $x$ be a member of the set on the right-hand side.

$x \in (A \cap C) \backslash (B \cap C)$ .

$\implies x \in A \cap C$ and $x \not\in B \cap C$ .

Note that $x \not\in B \cap C$ is equivalent to:

$x \not \in B$ , OR
$x \not\in C$ , OR
both $x \not\in B$ AND $x \not \in C$ .

However, $x \in A \cap C$ implies that $x \in A$ AND $x \in C$ .

The fact that $x \in C$ means that the only possibility that $x \not\in B \cap C$ is $x \not \in B$ .

To reiterate: if $x \not \in C$ , then the assumption that $x \in A \cap C$ would not be true any more. Therefore, the only possibility is that $x \not \in B$ .

Therefore, $x \in (A \backslash B)\cap C$ .

In other words, $x \in (A \cap C) \backslash (B \cap C) \implies x \in (A \backslash B)\cap C$ .

$(A \cap C) \backslash (B \cap C) \subseteq (A \backslash B)\cap C$ .

Combine these two parts to obtain: $(A \backslash B) \cap C = (A \cap C) \backslash (B \cap C)$ .

8 0

3 years ago

A 2-column table with 7 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3. The sec

oee [108]

The true statements about the function are:

f(x) ≤ 0 over the interval [0, 2].
f(x)>0 over the interval (-2,0)
f(x) ≥ 0 over the interval [2,∞)

<h3>How to determine the true statement?</h3>

The table of values is given as:

x f(x)

-3 -15

-2 0

-1 3

0 0

1 -3

2 0

3 15

From the table, the function values do not exceed 0 at the interval 0 to 2.

This means that f(x) ≤ 0 over the interval [0, 2].

Also from the table, the function values exceeds 0 at the intervals greater than -2 but less than 0

This means that f(x)>0 over the interval (-2,0)

Also from the table, the function values are at least 0 at the intervals greater than 2

This means that f(x) ≥ 0 over the interval [2,∞)

Hence, the true statements are (b), (d) and (e)